Hint: Log(xy)=log x + log y and If we have log a, a must be >0
This is how I would proceed:
We rely on the principle that log(xy) = log(x) + log(y).
LHS means left-hand-side, RHS right-hand-side.
LHS: log(x+3) + log(x+4) = log((x+3)(x+4)) = log(x^2 + 7x + 12)
Since this is the same as the RHS, all we need to do is ensure that the quantities (x+3), (x+4) and (x^2 + 7x + 12) are positive -- otherwise, taking their log is undefined.
So we are solving the simultaneous inequalities
x + 3 > 0
x + 4 > 0
(x+3)(x+4) > 0